Roboguru
SD

Diketahui Ksp​ X(OH)2​ dalam air adalah 3,2×10−11. Larutan jenuh  akan mengendap pada pH sebesar ....

Pertanyaan

Diketahui begin mathsize 14px style K subscript sp end style begin mathsize 14px style X open parentheses O H close parentheses subscript 2 end style dalam air adalah begin mathsize 14px style 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style. Larutan jenuh begin mathsize 14px style X open parentheses O H close parentheses subscript 2 end style akan mengendap pada pH sebesar ....space 

  1. 12 + log 4space 

  2. 10 + log 4space 

  3. 9 + log 2space 

  4. 8 + log 4space 

  5. 7 + log 2space 

S. Ibabas

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.space 

Pembahasan

Larutan akan mengendap apabila nilai begin mathsize 14px style Q subscript sp end style>begin mathsize 14px style K subscript sp end style. Nilai begin mathsize 14px style K subscript sp end style begin mathsize 14px style X open parentheses O H close parentheses subscript 2 end style = begin mathsize 14px style 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style

Langkah pertama yaitu menentukan kelarutan (s) dari begin mathsize 14px style K subscript sp end style
 

begin mathsize 14px style X open parentheses O H close parentheses subscript 2 equilibrium X to the power of 2 plus sign and 2 O H to the power of minus sign space space s space space space space space space space space space space space space space s space space space space space space space space space 2 s end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp end cell equals cell open square brackets X to the power of 2 plus sign close square brackets open square brackets O H to the power of minus sign close square brackets squared end cell row cell K subscript sp end cell equals cell s middle dot left parenthesis 2 s right parenthesis squared K subscript sp end subscript equals 4 s cubed end cell row s equals cell cube root of K subscript sp over 4 end root end cell row s equals cell cube root of fraction numerator 3 comma 2 cross times 10 to the power of negative sign 11 end exponent over denominator 4 end fraction end root end cell row s equals cell cube root of 0 comma 8 cross times 10 to the power of negative sign 11 end exponent end root end cell row s equals cell cube root of 8 cross times 10 to the power of negative sign 12 end exponent end root end cell row s equals cell 2 cross times 10 to the power of negative sign 4 end exponent end cell end table end style 
 

Langkah selanjutnya adalah menentukan pH sebagai berikut.
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 s end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 open parentheses 2 cross times 10 to the power of negative sign 4 end exponent close parentheses end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 4 cross times 10 to the power of negative sign 4 end exponent end cell row blank blank blank row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space 4 cross times 10 to the power of negative sign 4 end exponent end cell row pOH equals cell 4 minus sign log space 4 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign left parenthesis 4 minus sign log space 4 right parenthesis end cell row pH equals cell 10 plus log space 4 end cell end table end style 
 

Jadi, jawaban yang benar adalah B.space 

14

5.0 (1 rating)

Rahma Husodo

Pembahasan lengkap banget Ini yang aku cari! Mudah dimengerti Bantu banget Makasih ❤️

Pertanyaan serupa

Diketahui Ksp​ Cr(OH)3​ dalam air adalah . Larutan jenuh  akan mengendap pada pH sebesar ....

61

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia